To start with, let's notice that we might just as well prove this fact for the cosines of rational multiples of $\pi$ as the sines. We just need to use the identity $\sin x=\cos\left(\frac{\pi}{2} - x\right)$, which gives us:
$$\sin \frac{a\pi}{b}=\cos\left(\frac{\pi}{2} - \frac{a\pi}{b}\right)=\cos\left(\frac{(b-2a)\pi}{2b}\right)$$
As this is the cosine of a rational multiple of $\pi$, we get the result for sines more or less for free from the result for cosines.
Once this preliminary is out of the way, the key thing to notice is that $\cos(nx)$ can always be written as a polynomial function of $\cos x$. For example:
\begin{align}
\cos 2x &= 2 \cos^2 x - 1 \\
\cos 3x &= 4 \cos^3 x - 3 \cos x\\
\cos 4x &= 8 \cos^4 x - 8 \cos^2 x + 1 \\
\cos 5x &= 16 \cos^5 x - 20 \cos^3 x + 5 \cos x
\end{align}
and so on. (Formulas taken from here.)
Why does this matter? It means that we can find an explicit polynomial which has $\cos \frac{a\pi}{b}$ as a root. To see an example of this, let's look at $\alpha = \cos \frac{2\pi}{5}$. Using the last formula above, we know that $$
\cos (2\pi)=\cos\left(5 \cdot \frac{2\pi}{5}\right)=16 \cos^5 \frac{2\pi}{5}- 20 \cos^3 \frac{2\pi}{5} + 5 \cos \frac{2\pi}{5}=16\alpha^5-20\alpha^3+5\alpha
$$
Since $\cos(2\pi)=1$, we know that $\alpha$ is a solution to the equation $16x^5-20x^3+5x=1$, or, to put it another way, a root of the polynomial $16x^5-20x^3+5x-1$. So, by definition, $\alpha$ is algebraic (i.e., not transcendental).
You can use multiple-angle formulas to do this in general. If you want to prove that $\beta=\cos\frac{a\pi}{b}$ is algebraic, you just need to use the formula for $\cos bx$. This will give you a polynomial expression for $\cos(a\pi)$ in terms of $\beta$; since we know that $\cos(a\pi)=\pm 1$, that's enough to prove that $\beta$ is algebraic.
So we just need to prove this key fact, which can be done by induction. For the base case, notice that $\cos x$ and $\cos 2x=2\cos^2 x -1$ are both polynomial functions of $\cos x$. For the inductive step, we'll suppose that $\cos nx$ and $\cos (n-1)x$ are polynomials, and use the identity given here for $\cos a + \cos b$, with $a=(n+1)x$, $b=(n-1)x$:
$$\cos(n+1)x + \cos(n-1)x=2\cos nx \cos x$$
This can be rearranged into the formula $\cos(n+1)x = 2 \cos x \cos nx - \cos(n-1)x$, which means that if $\cos nx$ and $\cos(n-1)x$ are polynomials in $\cos x$, then so is $\cos(n+1)x$. So the proof by induction is finished.
If you're interested in knowing more details about this kind of polynomial formula for $\cos nx$ in terms of $\cos x$, a useful keyword to search on would be "Chebyshev polynomials."
As a final note, we could try to do this proof directly for $\sin \frac{a\pi}{b}$, as there are multiple-angle formulas for $\sin$ which are similar to the ones for $\cos$. But it'd be a little bit trickier, because those formulas often involve both $\sin$ and $\cos$. For example, $\sin 2x=2\cos x \sin x$, and getting rid of the $\cos$ would involve taking a square root and doing a bunch of annoying case analysis involving signs.
I don't have much depth to offer here, but I do find the question quite engaging; so I fooled around with it for awhile and was able to affirm my hunch that, at the very least, the cosine of any rational multiple of $\pi$ is algebraic over $\Bbb Q$; the calcuations are pretty simple though a little "grungy"; here's what I've got:
Set
$\theta = \dfrac{p \pi}{q}, \; 0 \ne p, 0 < q \in \Bbb Z; \tag 1$
then
$e^{i\theta} = \cos \theta + i \sin \theta = \cos \left (\dfrac{p \pi}{q} \right ) + i \sin \left ( \dfrac{p \pi}{q} \right ); \tag 2$
$(e^{i\theta})^q = e^{iq \theta} = \cos q \theta + i \sin q \theta$
$= \cos q \left ( \dfrac{p \pi}{q} \right ) + i \sin q \left ( \dfrac {p \pi}{q} \right ) = \cos p \pi + i \sin p \pi = (-1)^p; \tag 3$
$(e^{i\theta})^q = (\cos \theta + i\sin \theta)^q = \displaystyle \sum_0^q i^{k}\dfrac{q!}{k!(q - k)!} \cos^{q - k} \theta \sin^k \theta; \tag 4$
we may break this sum into two sums, over even and odd $k$, which are then its real (even) and imaginary (odd) parts; we will effect this operation by introducing a new index $j$ such that
$\displaystyle \sum_0^q i^{k}\dfrac{q!}{k!(q - k)!} \cos^{q - k} \theta \sin^k \theta = \sum_{j = 0}^{ 2j \le q} i^{2j} \dfrac{q!}{(2j)!(q - 2j)!} \cos^{q - 2j} \theta \sin^{2j} \theta$
$+ \displaystyle \sum_{j = 0}^{ 2j + 1 \le q} i^{2j + 1} \dfrac{q!}{(2j + 1)!(q - 2j - 1)!} \cos^{q - 2j - 1} \theta \sin^{2j + 1} \theta; \tag 5$
since $i^{2j} = (-1)^j$ and $i^{2j + 1} = (-1)^j i$, the sums on the right may be re-written and we have
$(e^{i\theta})^q = \displaystyle \sum_{j = 0}^{ 2j \le q} (-1)^j \dfrac{q!}{(2j)!(q - 2j)!} \cos^{q - 2j} \theta \sin^{2j} \theta$
$+ \displaystyle i \sum_{j = 0}^{ 2j + 1 \le q} (-1)^j \dfrac{q!}{(2j + 1)!(q - 2j - 1)!} \cos^{q - 2j - 1} \theta \sin^{2j + 1} \theta; \tag 6$
we see from (3),
$(e^{i\theta})^q = (-1)^p, \tag 7$
that $(e^{i\theta})^q$ is real; therefore, the second sum on the right of (6) vanishes; being a real multiple of $i$, this entire sum represents the imaginary part of $(e^{i\theta})^q$; we therefore need not deal further with this second sum, and will from here on only focus on the first, which is the real part of $(e^{i\theta})^q$; thus,
$(e^{i\theta})^q = \displaystyle \sum_{j = 0}^{ 2j \le q} (-1)^j \dfrac{q!}{(2j)!(q - 2j)!} \cos^{q - 2j} \theta \sin^{2j} \theta = (-1)^p; \tag 8$
our next step is to use the identity
$\sin^2 \theta = 1 - \cos^2 \theta \tag 9$
to eliminate the $\sin^{2j} \theta$ factors in (8), replacing them by
$\sin^{2j} \theta = (\sin^2 \theta)^j = (1 - \cos^2 \theta)^j, \tag{10}$
whence
$(e^{i\theta})^q = \displaystyle \sum_{j = 0}^{ 2j \le q} (-1)^j \dfrac{q!}{(2j)!(q - 2j)!} (\cos^{q - 2j} \theta) (1 - \cos^2 \theta)^j = (-1)^p, \tag{11}$
or
$(e^{i\theta})^q - (-1)^p = \left (\displaystyle \sum_{j = 0}^{ 2j \le q} (-1)^j \dfrac{q!}{(2j)!(q - 2j)!} (\cos^{q - 2j} \theta) (1 - \cos^2 \theta)^j \right ) + (-1)^{p + 1} = 0, \tag{12}$
which shows that $\cos \theta = \cos (p\pi / q)$ satisfies the polynomial
$P_{p, \; q}[x] = \left (\displaystyle \sum_{j = 0}^{ 2j \le q} (-1)^j \dfrac{q!}{(2j)!(q - 2j)!} x^{q - 2j} (1 - x^2)^j \right ) + (-1)^{p +1} \in \Bbb Q[x], \tag{13}$
and is hence algebraic over $\Bbb Q$. In fact, it is easy to see that the coefficients of $P_{p, \; q}[x]$ are integers, so that indeed
$P_{p,\; q}[x] \in \Bbb Z[x]. \tag{14}$
Now it should come as no surprise that the "results" derived here are well-known, and deeply analyzed; see this wikipedia article on Chebyshev_polynomials.
Best Answer
The proof is correct. For a nit-pick, you should say you have a non-zero polynomial.
Note that it would even suffice to know that at least one of $\pi$ and $e$ is transcendental.